Let $f(x) = x^{2} + ax + b$, where $a,\ b \in R$. If $f(x) = 0$ has all its roots imaginary, then the roots of $f(x) + f^{'}(x) + f^{''}(x) = 0$ are

Question

Dr. Abubacker · Accepted Answer

(b)
Given, $f(x) = x^{2} - ax + b$ has imaginary roots
$	herefore$ Discriminant, $D < 0\ \ \  \Rightarrow \ a^{2} - 4b < 0$
Now, $f^{'}(x) = 2x + a$
$f^{''}(x) = 2$
Also, $f(x) + f^{'}(x) + f^{''}(x) = 0$ ...(i)
$\Rightarrow \ \ x^{2} + ax + b + 2x + a + 2 = 0$
$\Rightarrow \ \ \ \ \ x^{2} + (a + 2)x + b + a + 2 = 0$
$	herefore\ \ x = \frac{- (a + 2) \pm \sqrt{(a + 2)^{2} - 4(a + b + 2)}}{2}$
$= \frac{- (a + 2) \pm \sqrt{a^{2} - 4b - 4}}{2}$
Since, $a^{2} - 4b < 0$
$	herefore\ \ a^{2} - 4b - 4 < 0$
Hence, Eq. (i) has imaginary roots

Q. Let $f(x) = x^{2} + ax + b$, where $a,\ b \in R$. If $f(x) = 0$ has all its roots imaginary, then the roots of $f(x) + f^{'}(x) + f^{''}(x) = 0$ are

Explanation

Reviewed by Dr. Abubacker

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